You have found the following ages (in years) of all $4$ tigers at your local zoo: $ 22,\enspace 13,\enspace 18,\enspace 16$ What is the average age of the tigers at your zoo? What is the variance? Round your answers to the nearest tenth. Average age: $ $
Solution: Because we have data for all $4$ tigers at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$. To find the population mean, add up the values of all $4$ ages and divide by $4$. $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{22 + 13 + 18 + 16}{{4}} = {17.25\text{ years old}} $ Find the squared deviations from the mean for each tiger. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $22$ years $4.75$ years $22.56$ years $^2$ $13$ years $-4.25$ years $18.06$ years $^2$ $18$ years $0.75$ years $0.56$ years $^2$ $16$ years $-1.25$ years $1.56$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean, we can find the population variance $({\sigma^2})$, without introducing any bias, by simply averaging the squared deviations from the mean: $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{22.56} + {18.06} + {0.56} + {1.56}} {{4}} $ $ {\sigma^2} = \dfrac{{42.74}}{{4}} = {10.69\text{ years}^2} $ The average tiger at the zoo is $17.3$ years old. The population variance is $10.7$ years $^2$.